Let $L/K$ be an extension of number fields, and $\theta\in \mathcal{O}_L$ a primitive element for $L/K$, so that $L = K(\theta)$.

The conductor of $(L/K,\theta)$ is the ideal of $\mathcal{O}_L$: $$\mathfrak{F} := \{\alpha\in\mathcal{O}_L\;|\;\alpha\mathcal{O}_L\subset\mathcal{O}_K[\theta]\}$$ How are the prime divisors of $\mathfrak{F}$ related to the ramified primes? Are they contained in the ramified primes? Are they equal to the ramified primes? Is there an example where the conductor is divisible by a non-ramified prime?

I know in the case of conductors of elliptic curves over $\mathbb{Q}$, the primes dividing the conductor are exactly the primes of bad reduction. Is something like that true here too?


1 Answer 1


Write $D_{L/K}$ for the different. The key statement is that $(f’(\theta))=D_{L/K}\mathfrak{F}$ (see the proof of Theorem III.2.5 in Neukirch’s Algebraic Number theory).

That proof is mostly formal, the only nontrivial point is that an element $x \in L$ lies in $f’(\theta)^{-1}O_K[\theta]$ iff $Tr_{L/K}(xO_K[\theta])$ is contained in $O_K$.

This tells us that… anything can happen, really.

Consider a quadratic extension $K/\mathbb{Q}$ with $O_K=\mathbb{Z}\oplus\mathbb{Z}u$, and let $\theta=nu$ for some $n \geq 1$. Then $\mathfrak{F}=nO_K$, which may or may not be divisible by any ramified prime.

In an extension $L/K$ of $p$-adic fields, we can always write $O_L=O_K[\alpha]$ for some $\alpha$. Thus, if $P \subset O_L$ is a prime, $\theta$ is a good enough $P$-adic approximation of such an $\alpha$ (for $L_P/K_{P \cap O_K}$), $\mathfrak{F}$ will not be divisible by $P$.


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